VCE Chemistry Question Thread

[Syndicate]

hello doing a gas stoichiometry question
it says there are 45kg of octane & my step 1 is to find the mole of octane
which  I am doing by n = 45000/114.0
what is the correct no. of sig figs for my answer? does the 114.0 count? or do i go by the 45 kg?

You go by the least number of sig figs in the question. Therefore you would use 2 sig figs (45 kg) in this case.

[Butterflygirl]

Hi guys,

Why is it better to use a highly conductive aqueous gel containing KNO3 for a salt bridge compared to a strip of filter paper dipped in KNO3 solution?

What effect would it have on measurements taken from a galvanic cell?

I don't understand how conductivity works and how it relates to the salt bridge.


Also,
If copper wire is used as a salt bridge, the voltage generated is higher but how does this happen? because doesn't the copper only contain cations (Cu2+)? How does it balance the charges in the half-cells and increase the voltage compared to using KNO3 solution or something?


Thankyou so much

[ardria]

Hi there!

Why is this reaction not a redox reaction? My teacher said something about it not "being in order" but I'm unsure what she meant.

Cu_(s) + Ni²⁺_(aq) --> Cu²⁺_(aq) + Ni_(s)

[hello friends]

Hi there!

Why is this reaction not a redox reaction? My teacher said something about it not "being in order" but I'm unsure what she meant.

Cu_(s) + Ni²⁺_(aq) --> Cu²⁺_(aq) + Ni_(s)

Redox reactions only occur when the oxidant is above the reductant in the electrochemical series. In this case, the oxidant nickel is not above the reductant copper so a reaction won't occur.

[Syndicate]

Redox reactions only occur when the oxidant is above the reductant in the electrochemical series. In this case, nickel, the oxidant as it's being reduced, is not above copper, the reductant as it's being oxidized, so a reaction won't occur.

I should have checked the electrochemical series first  (the reaction would only work if it went backwards)

[LPadlan]

When zinc powder is sprinkled into an acidified solution of potassium dichromate, a reaction occurs that produces zinc ions and chromium(III) ions in solution.
a) write the oxidation half-equation for the reaction
b) write the reduction half-equation for the reaction
c) use your answers to parts a and b to write a balanced equation for the overall equation.

Please help, this question is really doing my head in. It is on page 121 unit 3&4

[Syndicate]

When zinc powder is sprinkled into an acidified solution of potassium dichromate, a reaction occurs that produces zinc ions and chromium(III) ions in solution.
a) write the oxidation half-equation for the reaction
b) write the reduction half-equation for the reaction
c) use your answers to parts a and b to write a balanced equation for the overall equation.

Please help, this question is really doing my head in. It is on page 121 unit 3&4

Zn(s) is going to Zn2+ (aq). Therefore it is oxidising, as there is an increase in its oxidation number.
a) Zn(s) -> Zn2+(aq) + 2e-

Cr2O7^2-(aq) is going to Cr3+. Although it is increasing in oxidation number, it is losing its oxygen atoms. Therefore it is reducing (Potassium is being neglected as we tend to write an ionic equation - potassium is in aq state before and after the reaction).

b) Cr2O7^2- (aq) + 14H+ (aq) + 6e- -> 2Cr3+(aq) + 7H2O (l)

c) Multiply the oxidising reaction by 3, in order to cancel out the electrons.
Cr2O7^2-(aq) + 3Zn(s) + 14H+ (aq) -> 2Cr3+ (aq) + 7H2O (l) + 3Zn2+ (aq)

[LPadlan]

Cr2O7^2-(aq) is going to Cr3+. Although it is increasing in oxidation number, it is losing its oxygen atoms. Therefore it is reducing (Potassium is being neglected as we tend to write an ionic equation - potassium is in aq state before and after the reaction).

Hi can you please clarify why potassium isn't included?

[Syndicate]

Hi can you please clarify why potassium isn't included?

Well, if I were to write the full reduction equation, it would look something like:


As K+ is the only ion that remains in an aqueous state at both sides of the equation, it can be considered a spectator ion, which is why it is omitted from the equation.

Edit: grammar fix 

[Gogo14]

With galvanic cells, why is the anode negative and cathode positive. Isnt the anode losing electrons, so shouldn't it have a net positive charge?

[lzxnl]

With galvanic cells, why is the anode negative and cathode positive. Isnt the anode losing electrons, so shouldn't it have a net positive charge?

The anode is the side of oxidation, where electrons are leaving. As galvanic cells operate spontaneously, electrons are thus flowing to their 'preferred' destination. Would electrons want to leave the positive or negative terminal?

[lovebiology]

Hi I don't understand Q10 of the 2015 VCAA exam at all. like how should i even start a feul cell questions like this?
Can someone please help me?  Thanks

[deStudent]

For this question http://m.imgur.com/PWvGX9q

I got it correct since I just went with the first option that was correct in my mind. But I'm not entirely convinced how C and D cannot be also correct answers? Isn't ammonia also a gas? So it behaves like every other gas? D seemed dodgy since there are interparticle forces but they're negligible, right?

[Shadowxo]

For this question http://m.imgur.com/PWvGX9q

I got it correct since I just went with the first option that was correct in my mind. But I'm not entirely convinced how C and D cannot be also correct answers? Isn't ammonia also a gas? So it behaves like every other gas? D seemed dodgy since there are interparticle forces but they're negligible, right?

The key here is "exactly". Although you'd normally say for C and D that they're equal, they do have some (pretty negligible) interparticle forces of attraction, so they're not "exactly" the same (although only by a small amount). The larger molecule has more interparticle attraction so occupies a smaller volume. B is right because it's "approximately".

With galvanic cells, why is the anode negative and cathode positive. Isnt the anode losing electrons, so shouldn't it have a net positive charge?

I remember it as: the negative electrons are attracted to the positive cathode which is why they move.
You don't really need to know the specifics (I don't know the specifics that well), but even though the anode is "losing" electrons, there's a salt bridge so when the solution becomes more positive, negative molecules from the salt bridge eg NO3-, so overall it doesn't become more positive, it just has a lot of negative charge going through it (which is replaced as it leaves).

[Ahmad_A_1999]

Is there a way to write and balance the half equations in a fuel cell? Or will I always be given them?